A 1 00 g block of metal with a specific heat of 0.2 cal/gC degree that is at 100

Question

A 1 00 g block of metal with a specific heat of 0.2 cal/gC degree that is at 100 degree C is dropped into a large Styrofoam cup that contains 100 g of water that is at 20 degree C. Compare the heat lost/gained by the water and the metal block. A numerical value is not needed here, hut explain your response. Compare the temperature change of the water and die metal block. Which will change temperature the most? Numerical values are not needed at this point, hut explain your response. What is the final temperature of the mixture? Do a calculation and start with energy conservation: Q_H + Q_C = 0.

Explanation / Answer

Here ,

specific heat of metal , Sm = 0.2 Cal/(gm.degree C)

mass , m = 100 gm

a) here , as the styrofoam cup will insulate the water from the surronding

energy gained by water = energy lost by metal

b)

specific heat of water , Sw = 1 Cal/(gm.degree C)

let the temperature change of water is Tw

temperature change of metal is Tm

m * Sm * Tm = m * Sw * Tw

0.2 * Tm = 1 * Tw

Tm/Tw = 5

the temperature change of metal will be 5 times that of water

the metal will change the temperature the most

c)

let the final temperature is Tf

m * Sm * Tm = m * Sw * Tw

0.2 * (100 - T) = 1 * (T - 20)

solving for T

T = 33.33 degree C

the final temperature is 33.33 degree C

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